Answer:
9 + (-6)
Step-by-step explanation:
9-6=3
9 + (-6)= 3
Hope it will help you
Given:
ΔONP and ΔMNL.
To find:
The method and additional information that will prove ΔONP and ΔMNL similar by the AA similarity postulate?
Solution:
According to AA similarity postulate, two triangles are similar if their two corresponding angles are congruent.
In ΔONP and ΔMNL,
(Vertically opposite angles)
To prove ΔONP and ΔMNL similar by the AA similarity postulate, we need one more pair of corresponding congruent angles.
Using a rigid transformation, we can prove
Since two corresponding angles are congruent in ΔONP and ΔMNL, therefore,
(AA postulate)
Therefore, the correct option is A.
Its representedf by 2 polygons(triangles)
first one :
24-20=4
11-6=5
second:
24-14=10
11-7=4
B. frind area of 2 small trianges
a=(base x height )/2
first=4x5/2=10ft square
second=10x 4/2=20ft square
add them=10+20=30
rectangle24x 11=264
264-30=234
9514 1404 393
Answer:
1. ∠EDF = 104°
2. arc FG = 201°
3. ∠T = 60°
Step-by-step explanation:
There are a couple of angle relationships that are applicable to these problems.
- the angle where chords meet is half the sum of the measures of the intercepted arcs
- the angle where secants meet is half the difference of the measures of the intercepted arcs
The first of these applies to the first two problems.
1. ∠EDF = 1/2(arc EF + arc UG)
∠EDF = 1/2(147° +61°) = 1/2(208°)
∠EDF = 104°
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2. ∠FHG = 1/2(arc FG + arc ES)
128° = 1/2(arc FG +55°) . . . substitute given information
256° = arc FG +55° . . . . . . multiply by 2
201° = arc FG . . . . . . . . . subtract 55°
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3. For the purpose of this problem, a tangent is a special case of a secant in which both intersection points with the circle are the same point. The relation for secants still applies.
∠T = 1/2(arc FS -arc US)
∠T = 1/2(170° -50°) = 1/2(120°)
∠T = 60°
Answer:
The Standard Deviation is a measure of how spread out numbers are. You might like to read this simpler page on Standard Deviation first. But here we explain the formulas. The symbol for Standard Deviation is σ (the Greek letter sigma)
